Here is the step-by-step solution to each problem using the
Pythagorean Theorem and other relevant geometry formulas. Answers are given to
2 decimal places as instructed.
---
Problem 1: A4 Paper Diagonal
An A4 paper measures
210 mm by 297 mm.
#### a) Calculate the diagonal
This is a rectangle, so the diagonal forms the hypotenuse of a right triangle with legs 210 mm and 297 mm.
Use the Pythagorean Theorem:
\[
d = \sqrt{a^2 + b^2} = \sqrt{210^2 + 297^2}
\]
Calculate:
- \(210^2 = 44100\)
- \(297^2 = 88209\)
- Sum = \(44100 + 88209 = 132309\)
- \(\sqrt{132309} \approx 363.74\)
✔ Answer: 363.74 mm
#### b) Measure the diagonal to check
*(This is a hands-on activity — you’d physically measure an A4 sheet diagonally with a ruler. The calculated value should match closely, within measurement error.)*
---
Problem 2: Diagonal of a Square (side = 10 cm)
In a square, the diagonal splits it into two 45-45-90 right triangles.
Use Pythagorean Theorem:
\[
d = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} \approx 14.14
\]
✔ Answer: 14.14 cm
*(Alternatively, for any square: diagonal = side × √2 → 10 × 1.4142 ≈ 14.14)*
---
Problem 3: Area of Isosceles Triangle (sides 9 cm, 9 cm, 8 cm)
We can use the
Pythagorean Theorem to find the height, then use area formula.
Let’s draw the height from the apex (between the two equal sides) to the base (8 cm). This height bisects the base into two segments of 4 cm each.
Now we have a right triangle with:
- Hypotenuse = 9 cm (one of the equal sides)
- Base = 4 cm
- Height = ? (let’s call it *h*)
Apply Pythagorean Theorem:
\[
h = \sqrt{9^2 - 4^2} = \sqrt{81 - 16} = \sqrt{65} \approx 8.06
\]
Now, area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\[
\text{Area} = \frac{1}{2} \times 8 \times 8.06 = 4 \times 8.06 = 32.24
\]
✔ Answer: 32.24 cm²
---
Problem 4: Area of Regular Hexagon (side = 6 cm)
A regular hexagon can be divided into
6 equilateral triangles, each with side length 6 cm.
Area of one equilateral triangle:
\[
A = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}
\]
Total area = \(6 \times 9\sqrt{3} = 54\sqrt{3}\)
Now compute numerically:
\(\sqrt{3} \approx 1.73205\)
\[
54 \times 1.73205 \approx 93.53
\]
✔ Answer: 93.53 cm²
*(Alternative approach: Use formula for regular hexagon area = \(\frac{3\sqrt{3}}{2} s^2\) → same result.)*
---
Problem 5: Ladder Against Wall
Ladder = 10 m (hypotenuse), distance from wall = 2 m (one leg), height up wall = ? (other leg)
Use Pythagorean Theorem:
\[
h = \sqrt{10^2 - 2^2} = \sqrt{100 - 4} = \sqrt{96} \approx 9.80
\]
✔ Answer: 9.80 m
---
##
✔ Final Answers Summary:
1. a)
363.74 mm
b) *(Measure physically — should match ~363.74 mm)*
2.
14.14 cm
3.
32.24 cm²
4.
93.53 cm²
5.
9.80 m
All answers rounded to
2 decimal places as required.
Parent Tip: Review the logic above to help your child master the concept of solving pythagorean theorem worksheet.
